# How many grams of magnesium phosphate (Mg3(PO4)2) will be produced from the reaction of magnesium with 54.21g of phosphoric acid (H3PO4) ?

Apr 27, 2018

$0.3305 \text{mol}$

#### Explanation:

First we need to write the balanced equation.
$3 M g + 2 {H}_{3} P {O}_{4} \rightarrow M {g}_{3} {\left(P O 4\right)}_{2} + 3 {H}_{2}$

Now we need to find how many moles are in 54.21g of phosphoric acid.
$54.21 \text{g" * "1mol"/"82.00g" = 0.6611"mol}$

Now that we have the moles, we can check how much magnesium phosphate is produced.

$0.6611 \text{mol" * "1mol"_(Mg_3(PO4)_2)/"2mol" = 0.3305"mol}$